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Why does argparse give me a list-in-a-list?

开发者 https://www.devze.com 2023-02-14 07:21 出处:网络
I just noticed a behavior in argparse that puzzled me (guess I\'d never used it for a dumb list of files before):

I just noticed a behavior in argparse that puzzled me (guess I'd never used it for a dumb list of files before):

import argparse
parser = argparse.ArgumentParser()
parser.add_argument('multi', action='append', nargs='+')
print(parser.parse_args())

This gives me the output:

~$ ./testargs.py foo bar baz
Namespace(multi=[['foo', 'bar', 'baz']])
~$ 

I expected multi to be ['foo', 'bar', 'baz'], not a list within a list. As-is, I'll have to grab args.multi[0] before processing, which isn't a big deal, but feels like an ugly wart, and I'd like to understand why it's there.

Am I doing something sil开发者_运维百科ly in add_argument, or is this just an unavoidable quirk?


You are calling

parser.add_argument('multi', action='append', nargs='+')

And it is taking all the arguments and appending as a single item in the multi list.

If you want it as individual items, just don't use append

parser.add_argument('multi', nargs='+')

From the docs

'append' - This stores a list, and appends each argument value to the list. This is useful to allow an option to be specified multiple times. Example usage:

>>> import argparse
>>> parser = argparse.ArgumentParser()
>>> parser.add_argument('--foo', action='append')
>>> parser.parse_args('--foo 1 --foo 2'.split())
Namespace(foo=['1', '2'])
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